If […] =2k , where m is a positive integer then which of the followin…

The correct answer is Option (3) → A, B and C only

Since \( (m+1)!=(m+1)\,m! \) and \( (m+2)!=(m+1)(m+2)\,m! \), the factors \( m! \), \( (m+1)! \) and \( (m+2)! \) come out of R₁, R₂ and R₃.

    \( \displaystyle \Delta=m!\,(m+1)!\,(m+2)!\begin{vmatrix}1&m+1&(m+1)(m+2)\\1&m+2&(m+2)(m+3)\\1&m+3&(m+3)(m+4)\end{vmatrix} \)

With \( R_{3}\to R_{3}-R_{2} \) and \( R_{2}\to R_{2}-R_{1} \), the determinant becomes

    \( \displaystyle \begin{vmatrix}1&m+1&(m+1)(m+2)\\0&1&2(m+2)\\0&1&2(m+3)\end{vmatrix} \)

    \( \displaystyle =2(m+3)-2(m+2) \)

    \( \displaystyle =2 \)

So \( \Delta=2\,m!\,(m+1)!\,(m+2)! \), and comparing with \( 2k \):

    \( \displaystyle k=m!\,(m+1)!\,(m+2)! \)

So k is the product \( m!\,(m+1)!\,(m+2)! \).

Checking the statements

(A) \( m! \) — Correct. It is one of the three factors of k.

(B) \( (m+1)! \) — Correct. It is one of the three factors of k.

(C) \( (m+2)! \) — Correct. It is one of the three factors of k.

(D) \( (m+3)! \) — Incorrect. It need not divide k: for m = 1, k = 12, but 4! = 24 does not divide 12.